Answer:
Example
0.5 mol of sodium hydroxide is dissolved in 2 dm3 of water. Calculate the concentration of the sodium hydroxide solution formed.
Concentration =
Concentration = 0.25 mol/dm3
Volume units
Volumes used in concentration calculations must be in dm3, not in cm3. It is useful to know that 1 dm3 = 1000 cm3. This means:
divide by 1000 to convert from cm3 to dm3
multiply by 1000 to convert from dm3 to cm3
For example, 250 cm3 is 0.25 dm3 (250 ÷ 1000). It is often easiest to convert from cm3 to dm3 before continuing with a concentration calculation.
Question
100 cm3 of dilute hydrochloric acid contains 0.02 mol of dissolved hydrogen chloride. Calculate the concentration of the acid in mol/dm3.
Reveal answer
Converting between units
The relative formula mass of the solute is used to convert between mol/dm3 and g/dm3:
to convert from mol/dm3 to g/dm3, multiply by the relative formula mass
to convert from g/dm3 to mol/dm3, divide by the relative formula mass
Remember: the molar mass is the Ar or Mr in grams per mol.
Example
Calculate the concentration of 0.1 mol/dm3 sodium hydroxide solution in g/dm3. (Mr of NaOH = 40)
Concentration = 0.1 × 40
= 4 g/dm3
Answer:
Explanation:
The relation between equilibrium constant and Ecell is given below .
E⁰cell = (RT / nF ) lnK , F is faraday constant T is 273 + 25 = 298 K
E⁰cell = 1.46 - 1.21 = .25 V
n = 2
Putting the values
.25 = (8.314 x 298 lnK) / (2 x 96485 )
lnK = 19.47
K = 2.85 x 10⁸
2 )
Change in free energy Δ G
Δ G ⁰ = nE⁰ F
n = 4
E⁰ = .4 + .83 = 1.23 V
Δ G ⁰= 4 x 1.23 x 96485
= 474706 J / mol
3 )
E⁰cell = (RT / nF ) lnK
n = 2
1.78 = 8.314 x 298 lnK / 2 x 96485
lnK = 138.638
K = 1.62 x 10⁶⁰
Answer: The primary consumers are zooplankton, corals, sponges, Atlantic blue tang, and queen conch.
Explanation:
There will be 89.865 grams of Ar. If you put it in sig figs, it will be rounded to 89.9g At.
